Problem: What is the slope of the line tangent to $f(x) = x^{2}+4x-4$ at $x = 0$ ?
Answer: The slope of the tangent line is $ \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{((x+\Delta x)^{2}+4(x+\Delta x)-4) - (x^{2}+4x-4)}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{(x^{2}+2x \Delta x+\Delta x^{2}+4(x+\Delta x)-4) - (x^{2}+4x-4)}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{x^{2}+2(x \Delta x)+\Delta x^{2}+4x+4(\Delta x)-4-x^{2}-4x+4}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{2(x \Delta x)+\Delta x^{2}+4(\Delta x)}{\Delta x}$ $ = \lim_{\Delta x \to 0} 2x+\Delta x+4$ $ = 2x+4$ $ = (2)(0)+4$ $ = 4$